旅行商问题

题目：旅行商问题

给 n 个城市(从 1 到 n)，城市和无向道路成本之间的关系为3元组 [A, B, C]（在城市 A 和城市 B 之间有一条路，成本是 C）我们需要从1开始找到的旅行所有城市的付出最小的成本。

一个城市只能通过一次。
你可以假设你可以到达所有的城市

示例：

输入: 
n = 3
tuple = [[1,2,1],[2,3,2],[1,3,3]]
输出: 3
说明：最短路是1->2->3

题解：DFS

public class Solution {
    class Edge {
        int end;
        int cost;

        public Edge(int end, int cost) {
            this.end = end;
            this.cost = cost;
        }
    }

    private int minCost;

    public void dfs(int city, Map<Integer, List<Edge>> mapper, int cityCount, int cost, int[] visited) {
        if (cityCount == visited.length - 1) {
            minCost = Math.min(minCost, cost);
        }
        //最优剪枝
        if (cost > minCost) {
            return;
        }

        for (Edge edge : mapper.get(city)) {
            if (visited[edge.end] == 1) {
                continue;
            }
            visited[edge.end] = 1;
            dfs(edge.end, mapper, cityCount + 1, cost + edge.cost, visited);
            visited[edge.end] = 0;
        }
    }

    public int minCost(int n, int[][] roads) {
        Map<Integer, List<Edge>> mapper = new HashMap<>();
        int[] visited = new int[n + 1];
        minCost = 100000000;
        if (n == 0) {
            return minCost;
        }

        //建图
        for (int i = 1; i <= n; i++) {
            mapper.put(i, new ArrayList<>());
        }
        for (int i = 0; i < roads.length; i++) {
            mapper.get(roads[i][0]).add(new Edge(roads[i][1], roads[i][2]));
            mapper.get(roads[i][1]).add(new Edge(roads[i][0], roads[i][2]));
        }

        visited[1] = 1;
        dfs(1, mapper, 1, 0, visited);
        return minCost;
    }
}

标签：旅行,end,int,minCost,DFS,cost,roads,visited,lintcode816
来源： https://www.cnblogs.com/greengages/p/15781851.html